You want to obtain a sample to estimate how much parents spend on their kids' birthday parties. Based on a previous study, you believe the population standard deviation is approximately [tex]\sigma = 59.1[/tex] dollars. You would like to be 99% confident that your estimate is within 2 dollar(s) of the average spending on birthday parties.

How many parents do you have to sample?

You need to sample at least 2,577 parents to estimate the average spending on kids' birthday parties with a 99% confidence level and a margin of error of 2 dollars.

Explanation:

To determine the required sample size, we can use the formula for sample size estimation:

n = (Z * σ / E)^2

Where:

n is the required sample size
Z is the z-score corresponding to the desired confidence level
σ is the population standard deviation
E is the margin of error

In this case, the desired confidence level is 99%, which corresponds to a z-score of approximately 2.576 (obtained from a standard normal distribution table).

The population standard deviation is given as σ = 59.1 dollars.

The margin of error is 2 dollars.

Substituting these values into the formula:

n = (2.576 * 59.1 / 2)^2

Simplifying the calculation:

n ≈ 2,576.16

Rounding up to the nearest whole number, we need to sample at least 2,577 parents to estimate the average spending on kids' birthday parties with a 99% confidence level and a margin of error of 2 dollars.

Learn more about sample size estimation here:

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